Three-Winding Transformer — Star Equivalent, Powers and Efficiency: 22-Elec-B7 Power Systems Engineering, December 2013, Q3

Problem 3: Three-Winding Transformer — Star Equivalent, Powers and Efficiency (20 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Part (a) — five transformer types used in the electric power system.

Generator step-up (GSU) transformers — large two-winding units, typically delta on the generator side and grounded wye on the HV side, that lift generator terminal voltage (13.8–25 kV) to transmission voltage. The delta winding also isolates the machine from zero-sequence current.
Autotransformers — used to interconnect two transmission voltages of comparable magnitude (500/230 kV, 230/115 kV). Because part of the winding is common, the rating for a given throughput is much smaller than an equivalent two-winding unit; they nearly always carry a buried delta tertiary to stabilise the neutral and supply station service.
Distribution transformers — pole-mounted or pad-mounted units stepping the primary feeder (25 kV, 14.4 kV) down to utilisation voltage (600/347 V, 240/120 V), designed for very low no-load loss because they are energised continuously at light load.
Instrument transformers — current transformers (CTs) and voltage/potential transformers (VTs, CVTs) that scale primary quantities down to the 5 A / 120 V metering and relaying range while isolating the secondary circuits from system voltage.
Regulating and phase-shifting transformers — on-load tap changers for voltage magnitude control, and quadrature-boost (phase-angle regulating) transformers that inject a series voltage in quadrature to control real power flow on parallel paths.

Other legitimate answers include grounding (zig-zag) transformers, which create a zero-sequence source on an ungrounded system, and converter/rectifier duty transformers for HVDC terminals and industrial loads.

Given. The star equivalent circuit of a three-winding transformer, all impedances already referred to the primary side, with the secondary voltage taken as the reference phasor.

Given data
Quantity	Symbol	Value
Primary branch impedance	$Z_p$	$0.01+j0.08\ \Omega$
Secondary branch impedance	$Z_s$	$0.01+j0.08\ \Omega$
Tertiary branch impedance	$Z_t$	$0.01+j0.08\ \Omega$
Secondary voltage (reference)	$V_2$	$440\angle0^\circ$ V
Secondary current	$I_2$	$100\angle-25^\circ$ A
Tertiary current	$I_3$	$80\angle-35^\circ$ A

Find. The star-point voltage $V_0$, the primary current and voltage, the tertiary voltage, the apparent power and power factor at all three terminals, and the efficiency.

[Figure not reproduced: The three-winding transformer of Figure (1) redrawn as its star equivalent. All three branch impedances meet at the internal node V₀; the leakage drops are taken along each leg in turn. See the official exam paper.]

Approach. Walk the star circuit from the known secondary terminal inwards to the star point, apply Kirchhoff's current law at that node to get the primary current, then walk outwards along the primary and tertiary legs; the powers follow from $S=VI^{*}$ at each terminal.

Convert the two known currents to rectangular form. Every subsequent step is an addition of complex numbers, so this is done once: $$I_2=100\angle-25^\circ=90.6308-j42.2618\ \text{A},\ I_3=80\angle-35^\circ=65.5322-j45.8861\ \text{A}.$$
Part (a) — walk from the secondary terminal to the star point. The secondary current leaves the node through $Z_s$, so the node voltage is the terminal voltage plus that leakage drop: $$V_0=V_2+I_2Z_s = 440+(90.6308-j42.2618)(0.01+j0.08),$$ $$I_2Z_s = 4.2873+j6.8278\ \text{V},$$ $$\boxed{V_0=444.2873+j6.8278 = 444.34\angle0.880^\circ\ \text{V}}$$
Part (b) — apply KCL at the star point. The primary current supplies both loaded windings, so $$I_1=I_2+I_3=(90.6308+65.5322)-j(42.2618+45.8861),$$ $$\boxed{I_1=156.1629-j88.1479 = 179.32\angle-29.443^\circ\ \text{A}}$$ Note the magnitudes do not add: 100 A and 80 A give 179.3 A, not 180 A, because the two load currents are 10° apart.
Complete part (b) by adding the primary leakage drop. The primary current flows into the star point through $Z_p$, so the primary terminal must stand above the node by that drop: $$V_1=V_0+I_1Z_p,\ I_1Z_p=(156.1629-j88.1479)(0.01+j0.08)=8.6135+j11.6116,$$ $$\boxed{V_1=452.901+j18.439 = 453.28\angle2.331^\circ\ \text{V}}$$
Part (c) — walk outwards along the tertiary leg. The tertiary current leaves the node, so the tertiary terminal sits below the star point by its own leakage drop (everything here is already referred to the primary, which is what the question means by “referred to the primary side”): $$V_3=V_0-I_3Z_t,\ I_3Z_t=(65.5322-j45.8861)(0.01+j0.08)=4.3262+j4.7837,$$ $$\boxed{V_3=439.961+j2.044 = 439.97\angle0.266^\circ\ \text{V}}$$
Part (d) — form the complex power at each terminal. Using $S=VI^{*}$ with the terminal voltage and the terminal current at each port: $$S_1=V_1I_1^{*}=81\,283\angle31.775^\circ = 69\,100.9+j42\,801.8\ \text{VA},$$ $$S_2=V_2I_2^{*}=44\,000\angle25.000^\circ = 39\,877.5+j18\,595.2\ \text{VA},$$ $$S_3=V_3I_3^{*}=35\,197\angle35.266^\circ = 28\,737.8+j20\,322.1\ \text{VA}.$$
Read the power factors off the angles of those three phasors. The power factor is the cosine of the angle of $S$, and all three are lagging because every angle is positive: $$\boxed{\cos\varphi_1=0.8501,\ \cos\varphi_2=0.9063,\ \cos\varphi_3=0.8165\ \text{(all lagging)}}$$ The primary power factor is not the average of the other two: it is the power factor of the phasor sum of the two loads plus the reactive power consumed inside the three leakage reactances.
Part (e) — take the ratio of output to input real power. Only the real parts matter: $$\eta=\frac{P_2+P_3}{P_1}=\frac{39\,877.5+28\,737.8}{69\,100.9}=\frac{68\,615.3}{69\,100.9},$$ $$\boxed{\eta=0.99297 = 99.30\ \text{per cent}}$$
Check the loss balance independently. The 485.6 W deficit must be exactly the $I^{2}R$ dissipation in the three branch resistances, and it is: $$P_{\text{loss}}=(|I_1|^{2}+|I_2|^{2}+|I_3|^{2})(0.01)=(32\,156.9+10\,000+6\,400)(0.01)=485.57\ \text{W}.$$ The agreement to the last figure confirms every current magnitude. Note that this is a copper-loss-only efficiency: the star model carries no magnetising branch, so core loss is excluded and a real unit would test a little below 99.3 %.

Quantity	Symbol	Result
Star-point voltage	$V_0$	$444.34\angle0.880^\circ$ V
Primary current	$I_1$	$179.32\angle-29.443^\circ$ A
Primary voltage	$V_1$	$453.28\angle2.331^\circ$ V
Tertiary voltage (referred to primary)	$V_3$	$439.97\angle0.266^\circ$ V
Primary apparent power / pf	$S_1$	81.28 kVA, 0.8501 lagging
Secondary apparent power / pf	$S_2$	44.00 kVA, 0.9063 lagging
Tertiary apparent power / pf	$S_3$	35.20 kVA, 0.8165 lagging
Copper loss	$P_{\text{loss}}$	485.6 W
Efficiency	$\eta$	99.30 %

Topic. The star (wye) equivalent circuit of a three-winding transformer and its use for load-flow and efficiency calculations.

Key relations. $V_0=V_2+I_2Z_s$; $I_1=I_2+I_3$; $V_1=V_0+I_1Z_p$; $V_3=V_0-I_3Z_t$; $S=VI^{*}$; $\cos\varphi=\Re(S)/|S|$; $\eta=(P_2+P_3)/P_1$. The branch impedances themselves come from the three short-circuit tests via $Z_p=\tfrac12(Z_{ps}+Z_{pt}-Z_{st})$ and its cyclic permutations.

Why this works. A three-winding transformer has three independent short-circuit measurements ($Z_{ps}$, $Z_{pt}$, $Z_{st}$) and therefore exactly three degrees of freedom in its leakage description, which is precisely what a three-branch star supplies. The star point $V_0$ has no physical existence — it is a mathematical node representing the common mutual flux — but once all quantities are referred to a common base the circuit obeys ordinary Kirchhoff laws, so the problem reduces to walking the three legs. The direction convention is the only thing to get right: current into the node raises the terminal above $V_0$, current out of the node drops the terminal below it.

Common pitfall. Sign errors on the tertiary leg — writing $V_3=V_0+I_3Z_t$ gives 448.7 V instead of 440.0 V and quietly turns a load into a source. The second recurrent error is adding the current magnitudes at the star point (180 A rather than 179.3 A) and the third is quoting a power factor from the current angle alone, which is only correct at the secondary terminal where the voltage happens to be the reference phasor.

Source. El-Hawary, Electrical Power Systems: Design and Analysis, Ch. 4 (transformer models, three-winding equivalent); Grainger & Stevenson, Power System Analysis, §2.7 (three-winding transformers).

Question 3 of 7: Three-Winding Transformer — Star Equivalent, Powers and Efficiency

Problem 3: Three-Winding Transformer — Star Equivalent, Powers and Efficiency (20 points)